CHAPTER PROBLEM Did Mendel s results from plant hybridization experiments contradict his theory? Gregor Mendel conducted original experiments to study the genetic traits of pea plants. In 1865 he wrote Experiments in Plant Hybridization, which was published in Proceedings of the Natural History Society. Mendel presented a theory that when there are two inheritable traits, one of them will be dominant and the other will be recessive. Each parent contributes one gene to an offspring and, depending on the combination of genes, that offspring could inherit the dominant trait of the recessive trait. Mendel conducted an experiment using pea plants. The pods of pea plants can be green or yellow. When one pea carrying a dominant green gene and a recessive yellow gene is crossed with another pea carrying the same green/yellow genes, the offspring can inherit any one of four combinations of genes, as shown in the table below. Because green is dominant and yellow is recessive, the offspring pod will be green if either of the two inherited genes is green. The offspring can have a yellow pod only if it inherits the yellow gene from each of the two parents. We can see from the table that when crossing two parents with the green/yellow pair of genes, we expect that ¾ of the offspring peas should have green pods. That is, P(green pod) = ¾. When Mendel conducted his famous hybridization experiments using parent pea plants with the green/yellow combination of genes, he obtained 580 offspring. According to Mendel s theory, ¾ of the offspring should have green pods, but the actual number of plants with green pods was 428. So the proportion of offspring with green pods to the total number of offspring is 428/580 = 0.738. Mendel expected a proportion of ¾=0.75, but his actual result is a proportion of 0.738. In this chapter we will consider the issue of whether the experimental results contradict the theoretical results, and, in so doing, we will lay a foundation for hypothesis testing, which is introduced in Chapter 8. Gene from Parent 1 Gene from Parent 2 Offspring Genes Color of Offspring Pod green green green/green green green yellow green/yellow green yellow green yellow/green green yellow yellow yellow/yellow yellow
ROUND-OFF RULE FOR 2 µ, σ, and σ Round results by carrying one more place than the number of decimal places used for the variable. If the values of are, round to one decimal place. IDENTIFYING UNUSUAL RESULTS WITH THE RANGE RULE OF THUMB The range rule of thumb may be helpful in the value of a. According to the of, most values should lie within standard deviations of the ; it is for a value to differ from the mean by than standard deviations. Maximum usual value = + Minimum usual value = - IDENTIFYING UNUSUAL RESULTS WITH PROBABILITIES x successes among n trials is an unusually high number of successes if the of or more is unlikely with a probability of or.
x successes among n trials is an unusually low number of successes if the of or fewer is unlikely with a probability of or. RARE EVENT RULE FOR INFERENTIAL STATISTICS If, under a given, the probability of a particular event is extremely small, we conclude that the is probably not. Example 2: Based on information from MRINetwork, some job applicants are required to have several interviews before a decision is made. The number of required interviews and the corresponding probabilities are: 1 (0.09); 2 (0.31); 3 (0.37); 4 (0.12); 5 (0.05); 6 (0.05). a. Does the given information describe a probability distribution? b. Assuming that a probability distribution is described, find its mean and standard deviation.
c. Use the range rule of thumb to identify the range of values for usual numbers of interviews. d. Is it unusual to have a decision after just one interview. Explain. EXPECTED VALUE The of a is the theoretical mean outcome for many trials. We can think of that mean as the in the sense that it is the that we would expect to get if the trials could continue. DEFINITION The expected value of a random variable is denoted by, and it represents the of the. It is obtained by finding the value of x P( x) E ( ) = x P x.
Example 3: There is a 0.9968 probability that a randomly selected 50-year old female lives through the year (based on data from the U.S. Department of Health and Human Services). A Fidelity life insurance company charges $226 for insuring that the female will live through the year. If she does not survive the year, the policy pays out $50,000 as a death benefit. a. From the perspective of the 50-year-old female, what are the values corresponding to the two events of surviving the year and not surviving? b. If a 50-year-old female purchases the policy, what is her expected value? c. Can the insurance company expect to make a profit from many such policies? Why? 5.3 BINOMIAL PROBABILITY DISTRIBUTIONS Key Concept In this section we focus on one particular category of : probability distributions. Because binomial probability distributions involve used with methods of discussed later in this book, it is important to understand
properties of this particular class of probability distributions. We will present a basic definition of a probability distribution along with, and methods for finding values. probability distributions allow us to deal with circumstances in which the belong to relevant, such as acceptable/defective or survived/. DEFINITION A binomial probability distribution results from a procedure that meets all of the following requirements: 1. The procedure has a of trials. 2. The trials must be. 3. Each trial must have all classified into (commonly referred to as and ). 4. The probability of a remains the in all trials.
NOTATION FOR BINOMIAL PROBABILITY DISTRIBUTIONS S and F (success and failure) denote the two possible categories of outcomes ( ) P S = p ( ) P F = 1 p = q n x p q ( ) P x Example 1: A psychology test consists of multiple-choice questions, each having four possible answers (a, b, c, and d), one of which is correct. Assume that you guess the answers to six questions. a. Use the multiplication rule to find the probability that the first two guesses are wrong and the last four guesses are correct.
b. Beginning with WWCCCC, make a complete list of the different possible arrangements of 2 wrong answers and 4 correct answers, then find the probability for each entry in the list. c. Based on the preceding results, what is the probability of getting exactly 4 correct answers when 6 guesses are made? BINOMIAL PROBABILITY FORMULA n! x n x P ( x) = p q for x = 0,1, 2,, n n x! x! where ( ) n x p q
The factorial symbol, denotes the of decreasing. Example 2: Assuming the probability of a pea having a green pod is 0.75, use the binomial probability formula to find the probability of getting exactly 4 peas with green pods when 5 offspring peas are generated.
5.4 MEAN, VARIANCE, AND STANDARD DEVIATION FOR THE BINOMIAL DISTRIBUTION Key Concept In this section, we consider important of a distribution, including,, and. That is, given a particular binomial probability distribution, we can find its,, and. A distribution is a type of, so we could use the formulas from 5.2. However, it is easier to use the following formulas: Any Discrete pdf Binomial Distributions 1. µ = x P( x) 1. µ = np 2 2 2. σ = ( x µ ) P( x) 3. x P( x) σ 2 = 2 µ 2 2. 2 σ = npq 4. x P( x) σ = 2 µ 2 3. σ = npq
RANGE RULE OF THUMB Maximum usual value: Minimum usual value: Example 1: Mars, Inc. claims that 24% of its M&M plain candies are blue. A sample of 100 M&Ms is randomly selected. a. Find the mean and standard deviation for the numbers of blue M&Ms in such groups of 100. b. Data Set 18 in Appendix B consists of 100 M&Ms in which 27 are blue. Is this result unusual? Does it seem that the claimed rate of 24% is wrong?
Example 2: In a study of 420,095 cell phone users in Denmark, it was found that 135 developed cancer of the brain or nervous system. If we assume that the use of cell phones has no effect on developing such cancer, then the probability of a person having such a cancer is 0.000340. a. Assuming that cell phones have no effect on developing cancer, find the mean and standard deviation for the numbers of people in groups of 420,095 that can be expected to have cancer of the brain or nervous system. b. Based on the results from part (a), is it unusual to find that among 420,095 people, there are 135 cases of cancer of the brain or nervous system? Why or why not? c. What do these results suggest about the publicized concern that cell phones are a health danger because they increase the risk of cancer of the brain or nervous system?